Ideal Gas Equation And Gas Laws

Question

A sealed rigid container holds an ideal gas whose absolute temperature is doubled while its volume is kept fixed, so what happens to the pressure exerted on the walls?

RankGuru · Accepted Answer

The behaviour here is governed by the ideal gas equation from NCERT Class 11, Chapter 13 (Kinetic Theory), namely $PV = nRT$, which links pressure, volume and absolute temperature for a fixed amount of gas. Since the container is rigid and sealed, both the volume $V$ and the number of moles $n$ stay constant, so pressure is directly proportional to absolute temperature, giving the Gay-Lussac relation $P \propto T$. Writing $\frac{P_1}{T_1} = \frac{P_2}{T_2}$ and substituting $T_2 = 2T_1$ yields $P_2 = 2P_1$, so the pressure exactly doubles. The option claiming pressure halves reverses the proportionality and would apply if temperature were halved. The option that pressure is unchanged ignores that heating increases molecular collision frequency and momentum transfer to the walls. The option that pressure quadruples wrongly assumes $P \propto T^2$, which no gas law supports. A quick plausibility check confirms the answer: temperature is measured in kelvin, doubling a positive absolute temperature gives a positive doubled pressure, consistent in both sign and magnitude with a heated rigid vessel.

Ideal Gas Equation And Gas Laws

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